The ten digits 0,1,2,3,...9, are arranged at random with no repeats. What is the probability that the number thus formed represents
a.)a # greater than 6 billion?
b.) an even # greater than 6 billion?(HINT- There are two cases to consider: 1st digit odd and 1st digit even)

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The ten digits 0,1,2,3,...9, are arranged at random with no repeats. What is the probability that the number thus formed represents
a.)a # greater than 6 billion?
b.) an even # greater than 6 billion?(HINT- There are two cases to consider: 1st digit odd and 1st digit even)

Mathematics

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KingGeorge

6 years ago

Well, for a, first we have to find how many numbers above 6 billion we can make. As it turns out, if we have a first digit of 6 or greater, the number will be greater than 6 billion. So the first digit can be 6, 7, 8, or 9. That leaves us with 9 other digits.
This is a time for permutations. We have 9 spaces, and 9 digits, so have \(9!\) ways to represent the last 9 digits. We also have 4 possible numbers for the first digit, so we have \[4\cdot9!\]ways to make a number greater than 6 billion.

KingGeorge

6 years ago

For b, it's more complex.
Case 1: Even digit in front
This means that you either have a 6, or an 8 in front. Either way, you have 4 other even digits left. Once of these must go at the end. Then you have 8 digits left to fill in the rest of the places. So your answer should be \[2\cdot8!\cdot4\]Two choices for the first digit, 8! choices for the middle 8 digits, and 4 choices for the last digit.
Case 2: Odd digit in front
Now you have a 7 or a 9 in the front. This means that you have 5 even digits to choose from to use for the last digit. So your answer would be \[2\cdot8!\cdot5\]Two choices for the first digit, 8! choices for the middle 8 digits, and 5 choices for the last digit.

anonymous

6 years ago

OH! I see, so this is like those problems where you put all those blanks and fill them with numbers like 4.9.8.7.6.5.4.3.2.1? (The periods are the multiplication signs)

More answers

ya, I like it when it's a bit more visual so I can see how things work out. So what happens next to the 2.8!.5 and 2.8!.4?

KingGeorge

6 years ago

What do you mean by "happens next to ..."?

anonymous

6 years ago

Do you do anything else with them, or is that the end of the problem?

anonymous

6 years ago

When they become fractions for the probablity, how do you do that?

KingGeorge

6 years ago

Well, since it's asking for the total probability of the number being odd, you would have to add them together next. So you get \[(2\cdot8!\cdot4)+(2\cdot8!\cdot5)=2\cdot8!\cdot(4+5)=2\cdot8!\cdot9=2\cdot9!\]Then to find the probability, divide this number by the total amount of numbers above 6 billion.\[{2\cdot9! \over 4\cdot9!}={1 \over 2}\]

anonymous

6 years ago

The answer sheet says that it is 1 over 5... hmmm... very peculiar...

KingGeorge

6 years ago

Ah. I see what it was asking for the probability of now. Instead of dividing by \(4\cdot9!\) we need to divide by the total amount of numbers you can create. Since you have 10 digits, and 10 places, you can make \(10!\) total numbers. So the probability is \[{2\cdot9! \over 10\cdot9!}={2 \over 10}={1 \over 5}\]

KingGeorge

6 years ago

Likewise, for a, you want \[{4\cdot9! \over10\cdot9!}={2\over5}\]

anonymous

6 years ago

Thank you so very much KingGeorge. You are a god!
I know that this will help me tons on the math quiz I have on this tomorrow!!! Wish me lots of luck!!!Actually, that's an insult to math people.Wish me lots of ...skill and... logic!!!

KingGeorge

6 years ago

Good skill and logic! With maybe just a little bit of luck thrown in there for good measure.

anonymous

6 years ago

hey can anyone help me too with functions???...:(
i'm having a problem in it...

anonymous

6 years ago

Haha! Thank you so much! I appreciate it more than you would know!!! Bye, I got to go to sleep so I can ace that quiz tomorrow! Good skill and Logic to you too!!!